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Math Help - Polar Graph question..

  1. #1
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    Polar Graph question..

    Consider r= cos2theta in polar form.
    I understand that if you want the area of one loop you would integrate ( o.5r^2) between -pi/4 to pi/4.

    But what is the interpretation of integrating between -pi to pi?
    I know it is something to do with r becoming negative but what am i working out in this case?
    Also do negative r areas cancel out positive r areas like they do in cartesian?
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  2. #2
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    Quote Originally Posted by rodders View Post
    Consider r= cos2theta in polar form.
    I understand that if you want the area of one loop you would integrate ( o.5r^2) between -pi/4 to pi/4.

    But what is the interpretation of integrating between -pi to pi? ... the area of all four petals

    Also do negative r areas cancel out positive r areas like they do in cartesian? they can, when you get an "inner loop" ... graph the cardioid r = 0.5 + cos(t) to see what I'm talking about.
    ...
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  3. #3
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    How can areas cancel each other out when you are essentially integrating something squared?
    In which case finding the area of the cardoid 0.5 + 1cos(theta)...you would be wrong to integrate between -pi and pi because of the inner loop?
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  4. #4
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    ^ Nobody got an ideas here? Its really puzzling me? Ican't see how the area can be negative for r^2 without messing with the limits? But why would you?So how can a inner loop have a negative area?
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  5. #5
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    My error ... the inner loop gets counted twice as a positive value, i.e. not negative. The only way I can see to get a negative area is to change the direction of rotation of the sweeping ray , i.e. from \pi to -\pi
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